I've had a few attempts now at coming up with a rigorous proof for the divisibility of $n!$ by the product of factorials in the $k^{th}$ multinomial coefficient in $m$ variables, the statement below is what I "feel" as if I should be stating, but there seem to be obvious steps and assertions missing at least until I consider it to be vigorous, let alone everyone else.
The combined conditions of:
1) these factorials in the product being equal to the exponents of the variable product in the $k^{th}$ monomial (or individual term in the multinomial expansion if you like)
2) Their values being positive integers between 0 and $n$
3) The sum of the arguments of the factorials must be equal to $n$.
Seem to be the best justification I can think of at this point as to why we are always guaranteed the assurance of the divisibility relation being true, so I have stated this as follows: $$\sum _{j=1}^{m}\alpha_{{k,\,j}}=n\, \land \,0\leq \alpha_{{k,\,j}}\leq n \Rightarrow \prod _{j=1}^{m}\alpha_{{k,j}}!\,\, | \,n!$$
Edit: Kummer's Theorem may be generalized to apply to the multinomial coefficients and prove this divisibility relation. wiki
Seems like you've already noticed that you can prove this by looking at the powers of primes dividing the factorials (essentially your generalization of Kummer's theorem), but if you prefer a combinatorial approach,
$$\frac{n!}{\prod_{j=1}^{m}\alpha_{j}!}$$
(for $\alpha_j$ which sum to $n$) counts the number of different permutations of $\alpha_1$ 1's, $\alpha_2$ 2's, ..., $\alpha_m$ m's. This is because there are $n!$ ways to arrange all $n$ symbols, but this overcounts by a factor of $\prod \alpha_j!$, since there are $\alpha_j!$ ways to rearrange the set of $\alpha_j$ $j$s.